anonymous
  • anonymous
in a particular game a player throws two fair dice together. if their total is 7 or 11 he wins if their total is 2,3 or 12 he looses for any other total, he neither wins or looses. (i) find the probability of winning (ii) find the probability of losing
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Start with listing all your possible dice pair outcomes and count the number of 7's and 11's out of all possible outcomes. That gives your probability of winning. Same process with the losing for 2, 3, and 12.
anonymous
  • anonymous
So start with "all possible outcomes". How many of those are there? hint: each die has 6 faces.
anonymous
  • anonymous
am 2 dice so 12 ?

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anonymous
  • anonymous
Better start with one die. How many outcomes?
anonymous
  • anonymous
6 ?
anonymous
  • anonymous
Yes. Now for EACH GIVEN outcome on that first die, you can have 6 outcomes on the second die. So, 6 x 6 = 36 possible outcomes for a roll of 2 die. Now, it really would be beneficial for you to write out on paper a 2-dimensional table where horizontal is one die and vertical is the other die and make a table of 36 rolls. That will benefit you immensely here.
anonymous
  • anonymous
is the answer for part (i) 2over 9 ?
anonymous
  • anonymous
Yes, 2/9 for winning. That comes from 6 7's and 2 11's inning rolls out of a possible 36: 8/36 = 2/9 Good job!
anonymous
  • anonymous
yeah ok thank you is is the same method for loosing aswell ?
anonymous
  • anonymous
Very similar. Add up all 2's (there's just 1), all 3's (there are 2) and all 12's (there's 1) for that total over 36.
anonymous
  • anonymous
1 over 9 ?
anonymous
  • anonymous
Yes, because 4/36 = 1/9. This would be a nice game to play because you win more than lose!
anonymous
  • anonymous
thank you :) i actuallt have a similar question to that and i was wondering if i got it right or not do you mind checking for me ?
anonymous
  • anonymous
Good luck in all of your studies and thx for the recognition! @samnatha I'll check it if you like.
anonymous
  • anonymous
thank you i think i sent to you in mail :)
anonymous
  • anonymous
i'll just put it here aswell :)
anonymous
  • anonymous
this circle is divided into 6 equal sectors. you pay €10 to spin the arrow and you win the amount in the sector where the arrow stops. what is the expected amount you win or loose in this game ? theres a picture of a cirlce aswell divided into 6 sections in each section there is different numbers which are €0 €30 €10 €10 €0 and €5
anonymous
  • anonymous
Since you pay 10 to play, we are really talking about the "net" winning or losing, so you can actually re-do the 6 sectors as a "net" amount: -10, +20, 0, 0, -10, -5 You add those up and divide by 6 to get -5/6. Not a good game to play. Over time, you lose big.
anonymous
  • anonymous
ok i'm slighlt confused on how you did all that not great at probability
anonymous
  • anonymous
np. Each sector is equally likely, and for each sector, we can conceptually see ourselves handing over 10 (at first, to play) and then getting back some money (if any for that sector). So, for example, if we landed on "get 30", it's a net of 20 because we had to pay 10 up front. Your expected value is the net amount times the probability of landing in that sector. It is 1/6 times each sector and then all added up. Or you could just add them all up first: (1/6)(-10) + (1/6)(20) + (1/6)(0) + (1/6)(0) + (1/6)(-10) + (1/6)(-5) or what I did: (1/6)(-10 + 20 + 0 + 0 + -10 -5)
anonymous
  • anonymous
ooh ok thank you i am use to doing it the other way so that made more sense thanks alot :)
anonymous
  • anonymous
yw!

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